(Java) 返回当前字符串的新字符串版本，其中所有 >= 或 <= 给定字符 n 的字母都被删除

Question

Returns a new string version of the current string where all the letters either >= or <= the given char n, are removed.

The given letter may be az or AZ, inclusive. The letters to be removed are case insensitive. If 'more' is false, all letters less than or equal to n will be removed in the returned string. If 'more' is true, all letters greater than or equal to n will be removed in the returned string. If the current string is null, the method should return an empty string. If n is not a letter (and the current string is not null), the method should return an empty string.

Questions: Since the method receives char n, how do I apply for loop to each character? Do I use arrayOfArg and is it correct to use arrayOfArg[i] ++ to store letters more than the letter?

public String filterLetters(char n, boolean more) {
    char [] arrayOfArg = n.toCharArray();
    String myNewString = "";
    String removedChar = "";
    
    if (n == null) {
        return "";
    }
    
    for (int i = 0; i < Character.length; i++) {
        if (more == false) {
            //all letters less than or equal to n will be removed in the returned string
            if (Character.isLetter(i) == true) continue;
            removedChar = arrayOfArg[i] ++;
            myNewString = arrayOfArg.replace(removedChar,"");
                
        }
        if (more == true) {
            //all letters less than or equal to n will be removed in the returned string
            if (Character.isLetter(i) == true) continue;
            removedChar = arrayOfArg[i] -- ;
            myNewString = arrayOfArg.replace(removedChar,"");
        }
    }
    return myNewString;
        } 

Answer 1

This is a slight modification to the code you requested but the output works properly. I'm sure someone else has a better idea than me. This is just a starting place.

public static String filterLetters(String word, String n, boolean more) {
    String[] alphabet = new String[]{"a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"};
    int charValue = 0;
    //Find Numerical Value of Char n
    for (int i = 0; i < alphabet.length; i++) {
      if (alphabet[i].equalsIgnoreCase(n)) {
        charValue = i + 1;
      }
    }
    if (more) {
      for (int i = charValue; i < alphabet.length; i++) {
        word = word.replaceAll(alphabet[i],"");
      }
    } else {
      for (int i = charValue; i > 0; i--) {
        word = word.replaceAll(alphabet[i],"");
      }
    }

    
    return word;
  }

Answer 2

Use regex!

public String filterLetters(String str, char n, boolean more) {
    if (str == null || !Character.isLetter(n)) {
        return "";
    }
    String range = more ? n + "-z" : "a-" + n;
    return str.replaceAll("(?i)[" + range + "]", "");
}

This works by creating a regular expression as follows:

• if more is true: (?i)[nz]
• if more is false: (?i)[an]

where "n" is the actual char passed in.

(?i) flag turns on case insensitivity.

[an] is the range (inclusive) of characters from a to n .
[nz] is the range (inclusive) of characters from n to z .

Note: The necessary String parameter has been added to the method.